leetcode中等114二叉树展开为链表

Posted 2021-06-06 qq_40707462

思路一：递归先序遍历
中左右，即把左放在右的位置，再把右接在左的下面

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        if not root or (not root.left and not root.right):
            return 
        #处理左右子树
        self.flatten(root.left)
        self.flatten(root.right)
        #中左右---中，把左放在右，把右放在左下面
        tmp=root.right
        root.right=root.left
        root.left=None
        #找到新树右的末尾，把原右（tmp）接在后面
        while(root.right):
            root=root.right
        root.right=tmp

思路二：非递归

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        stack=[]
        while root or stack:
            if root:
                stack.append(root)
                root=root.left
            else:
                node=stack.pop()
                tmp=node.right
                root=tmp
                #以上是前序遍历，以下是每个结点处的操作
                node.right=node.left
                node.left=None
                while node.right:
                    node=node.right
                node.right=tmp